A man with normal lungs and an arterial PO2 of 40 mmHg takes an overdose of barbiturates that halves his alveolar ventilation wi
1 answer:
Answer:
Where x represent the value of interest on this case. And solving for the value of x we have:
So then the new arterial pressure needs to be now 50 mm Hg to mantain the original level.
And in order to find the concentration we can use a figure called the "O2
-CO2 diagram showing a ventilation-perfusion ratio line. " and when we use this graph to calculate the pressure of Co2 for PO2= 40 mmHg and for Po2=50 mm Hg we got and increase of 0.07 or 7%
So then the final answer for this case would be an increase of 7%
Explanation:
For this case we know that a man with normal lungs have an arterial Po2 os 40 mm Hg.
Then we know that this man take an overdose of barbiturates thats halves his aveolar ventilation without changing his metabolism
We also know that the respiratory exchange ratio is 0.8 or 8/10
And we want to find hor much does his inspired oxyden concentration % have to increased to return his avelolar Po2 to the original level.
On this case we can apply a proportion rule and we have this:
Where x represent the value of interest on this case. And solving for the value of x we have:
So then the new arterial pressure needs to be now 50 mm Hg to mantain the original level.
And in order to find the concentration we can use a figure called the "O2
-CO2 diagram showing a ventilation-perfusion ratio line. " and when we use this graph to calculate the pressure of Co2 for PO2= 40 mmHg and for Po2=50 mm Hg we got and increase of 0.07 or 7%
So then the final answer for this case would be an increase of 7%
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Answer: 0.9375 g
Explanation:
To calculate the number of moles for given molarity, we use the equation:
.....(1)
Molarity of solution = 0.75 M
Volume of solution = 25.0 mL = 0.025 L
Putting values in equation 1, we get:
According to stoichiometry :
2 moles of require = 1 mole of
Thus 0.01875 moles of will require=
of
Mass of
Thus 0.9375 g of is required to react with 25.0 ml of 0.75 M HCl
Answer:
Explanation:
To answer this question successfully, we need to remember that atoms are neutral species, since the number of protons, the positively charged particles, is equal to the number of electrons, the negatively charged particles. That said, we may firstly find an atom which has 3 electrons (and, as a result, 3 protons, as it should be neutral).
The number of protons is equal to the atomic number of an element. We firstly may have an atom with 3 protons and 3 electrons (atomic number of 3, this is Li).
Similarly, we may take the atomic number of 4, beryllium, and remove 1 electron from it. Upon removing an electron, it would become beryllium cation, .
We may use the same logic going forward and taking the atomic number of 5. This is boron. In this case, we need to remove 2 electrons to have a total of 3 electrons. Removal of 2 electrons would yield a +2-charged cation: .
Answer:
The equilibrium will shift left.
Explanation:
Hope this helps <3
